ln(y-1) = x^2 /2 + C1 Mr F says: Wrong. It is ln|y - 1| NOT ln(y - 1). There is a big difference that you should be aware of from basic calculus.

Mutliplied both sides by 2

ln(y-1) + x^2 = C Mr F says: If you multiply both sides by 2, and then subtract x^2, then surely you get 2 ln|y - 1| - x^2 = C. Although I don't know why you moved the x^2. It is better left as 2 ln|y - 1| = x^2 + C.

= 1.099 Mr F says: This has come out of nowhere and is meaningless without explanation.

Again do I need to find 'C' to answer this question or just leave it with at the integrated solution?

Thanks

Obviously you need to find the value of C, why do you think you were "given that when x = -, y = 4" ? I assume you are meant to get y as a function of x, so I don't know why you are giving a single number as an answer.

Sorry, but from what I can see your problem is mainly algebra not calculus. You would be well-advised to take much greater care with the algebra. And it wouldn't hurt to review your class notes and textbook on solving differential equations.